Minimum cost flows
Amaury Pouly
November 23, 2010
Contents
1 Introduction
1.1 Mathematical formulation . . . . . . . . . . .
1.2 Applications to graph theory . . . . . . . . .
1.2.1 Shortest path problem . . . . . . . . .
1.2.2 Maximum flow problem . . . . . . . .
1.3 Applications to real world problems . . . . .
1.3.1 Determining an optimal energy policy
1.3.2 Ship loading problem . . . . . . . . .
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2
2
2
3
3
3
3
5
2 Mathematical properties
2.1 Flow decomposition . . . . . . . . . . .
2.2 Removing nonzero lower bounds . . . .
2.3 Removing infinite capacities . . . . . . .
2.4 Removing negative costs . . . . . . . . .
2.5 Making the network strongly connected
2.6 Residual network . . . . . . . . . . . . .
2.7 Negative cycle optimality condition . . .
2.8 Reduced cost . . . . . . . . . . . . . . .
2.9 Reduced cost optimality condition . . .
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5
6
7
7
8
9
9
11
12
13
3 Algorithms
3.1 Finding a feasible flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Cycle-cancelling algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Successive shortest path algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
14
14
15
4 Conclusion
18
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1
1
Introduction
Everywhere we look in our lives, networks are apparent. From electrical and power networks to manufacturing
and distribution networks, there are infinite variations between the most physical and the most abstract
networks. In all these problem domains, we wish to move some entity (electricity, a consumer product, a
message) from one point to another in the underlying network, and to do so as efficiently as possible.
The minimum cost flow model is the most fundamental of all network flow problems. The problem is easy
to state: we wish to determine a least cost shipment of a commodity through a network in order to satisfy
demands at certain nodes from available supplies at other nodes. This model has a number of familiar and
less familiar applications: the distribution of a product from plants to warehouses, or from warehouses to
retailers; the routing of cars through a street network; or even the orthogonal drawing of a planar graph.
In this section, we present a mathematical formulation of the minimum cost flow problem and some of its
applications.
1.1
Mathematical formulation
Let G = (N, A) be a directed graph, which is allowed to have parallel arcs. We denote by N + (v) the outneighbourhood of v, and by N − (v) the in-neighbourhood of v (here, the neighbourhood is thought in terms
of edges). We associated to each arc e three quantities:
• the cost c(e) per unit of flow on that arc
• the capacity u(e) that denotes the maximum amount that can flow on the arc
• the lower bound l(e) that denotes the minimum amount that must flow on the arc
We also associate to each node v a number b(v) representing its supply/demand. If b(v) > 0, node v is a
surplus node; if b(v) < 0, node v is a demand node with a demand of −b(i); and if b(v) = 0, node v is
a transshipment node. The minimum cost flow problem can be formulated as an optimization problem as
follows. The decision variables of the problem are the arc flows, which are denoted by f (e) for each arc e ∈ A.
Minimize
X
c(e)f (e)
e∈A
subject to
X
e∈N + (v)
f (e) −
X
f (e) = b(v)
e∈N − (v)
l(e) 6 f (e) 6 u(e)
where
X
for all v ∈ V
for all e ∈ A
v(v) = 0. Furthermore, we usually require that all quantity be integers because for most applica-
v∈V
tions we can reach this assumption by multiplying all the quantities by a suitably large number. Moreover,
irrational numbers generally need to be approximated by rational ones to represent them on a computer. We
call the equality constraint on the flows the mass balance constraint.
1.2
Applications to graph theory
A number of special cases of the minimum cost flow happen to play a central role in graph theory and more
generally in applications of network flows. We list two of them and explain why there are special cases of our
general model. Furthermore, those two applications have a deep relation with the minimum cost flow problem
since it shares common aspects with both of them. Finally, a number of minimum cost flow algorithms heavily
rely on those problems in order to solve our general problem.
2
c(e)
x
10
12
y
t
4
v0
5
0
1
s
v1
−3
,
(10
0)
(12
,
1)
−1
t
(4, 0)
s
b(x)
(c(e), f (e))
x
v0
b(y)
y
(5, 0
(0,
1
)
)
,
(−3
0
v1
1)
0
Shortest path graph (shortest path in red)
Network flow graph (minimum cost flow in red/blue)
Figure 1: Example of a shortest path problem and its mapping to the minimum cost flow model
1.2.1
Shortest path problem
The shortest path problem is one of the simplest of all network flow problems. For this problem, we wish
to find a path of minimum cost (or length) from a specified source node s to another specified sink node
t, assuming that each arc has an associated cost c(e). In order to fit this model to our minimum cost flow
problem, we set b(s) = 1, b(t) = −1 and b(v) = 0 for all other nodes. We further set l(e) = 0 and u(e) = ∞
for each arc in the graph. Intuitively, the try to send one unit of flow from s to t, and to do so at the
minimum possible cost. One can then see and the unit of flow follows the shortest path. Figure 1 summarizes
the transformation on an example.
In the case where we are interested in the shortest path from a source s to every other node in the graph,
we set b(s) = n − 1 and b(v) = −1 for all other nodes, where n = |V |. We further set l(e) = 0 and u(e) = ∞.
Intuitively, we send n − 1 units of flow from s and since all nodes demand exactly one unit (since b(v) = −1),
we wish to send exactly one unit of flow to each node. One can then easily show that minimizing the total
cost of all paths is equivalent to minimizing the cost of each path. Furthermore, the units of flow follow the
shortest paths.
1.2.2
Maximum flow problem
The maximum flow problem is in a sense a complementary model to the shortest path problem. Instead of
sending an unbounded amount of flow along a path of minimum cost, we wish to send a maximum amount
of flow through the network regardless of the cost. Formally, the maximum flow problem seeks a feasible
solution that sends the maximum amount of flow from a specified source s to a specified sink t, given a
maximum bound on the amount we can send though an edge. We can formulate this problem as a minimum
cost flow problem in the following manner. We set b(v) = 0 for all v ∈ V , c(e) = 0 for all e ∈ A and keep the
maximum bound u(e). We further introduce an arc (t, s) with cost c(t, s) = −1 and a flow bound u(t, s) = ∞.
Then the minimum cost flow maximizes the flow on the arc (t, s); but since any flow on arc (t, s) must travel
from node s to t through the other arcs of network, we will maximize the flow from s to t in the original
network. Figure 2 summarizes the transformation on an example.
1.3
1.3.1
Applications to real world problems
Determining an optimal energy policy
Most countries or firms need to decide on an energy policy, i.e. how to utilize the available raw materials and
transformations to satisfy their energy needs. Consider, for example, a country which has four raw materials:
crude oil, coal, uranium and hydropower; and has four energy needs: electricity, domestic oil, petroleum and
gas. The country has the necessary technology to convert crude oil into domestic oil or petroleum, coal into
electricity and gas, and uranium and hydropower into electricity. Each available transformation comes at a
specified cost per unit and a maximum amount of transformed material per year. Finally, the country can
3
(u(e), f (e))
x
b(x)
x
y
(u(e), c(e), f (e))
v1
t
s
(3,
0, 3
)
5)
(5,
v0
(∞, −1, 7)
Network flow graph (maximum flow in red)
(2, 0, 0)
(2, 0)
)
0, 4
(4,
(7,
0, 2
)
(2, 0, 2)
(3,
3)
(2, 2)
s
v1
(7,
2)
4)
(4,
b(y)
y
)
0, 5
,
5
(
t
v0
Network flow graph (minimum cost flow in red/blue)
Figure 2: Example of a maximum flow problem and its mapping to the minimum cost flow model
Electricity
Crude oil
1
1
2
Coal
2
Domestic oil
s
t
3
Uranium
3
Petroleum
4
Hydropower
4
Gas
Figure 3: Energy policy problem formulated as a minimum cost flow problem
only produce a certain amount of each raw material at a specified cost per unit. The objective is to satisfy,
at the least possible cost, a certain annual consumption level of various energy needs.
Figure 3 shows the formulation of the problem as a minimum cost flow problem. The network has three
types of arcs:
• sources arcs e = (s, m) from the source s to a raw material m with the cost c(e) of producing the
material m, an upper bound u(e) on the production and a lower bound l(e) on the production (some
industry require a minimum production per year)
• conversion arcs e = (m, n) from a raw material m to an energy need n with a the cost c(e) of transformation per unit and upper bound u(e) on the amount of transformed material per year
• sink arcs e = (n, t) from a energy need n to the sink t with a cost c(e) = 0 and the lower bound l(e) on
the annual consumption of n
We further add an arc (t, s) with zero cost and unbounded capacity and set demand of nodes to zero:
b(v) = 0. Then the minimum cost flow problem is exactly the energy policy problem as stated below. Notice
that to formulate this problem in a more realistic way, we would need to introduce an efficiency value on
the conversion edges since one unit of a raw material does not necessary produces one unit of something else;
this is known as the generalized flow problem.
4
y
u(e)/c(e)
x
15
−f 15
14
−f 14
13
12
1
−f
5
−f 2 24
−f 13
23
12
r
2
25
−f
−f 24
34
23
r
35
3
−f
−f 35
34
r
4
45
−f
45
r
5
Figure 4: Ship loading problem formulated as a minimum cost flow problem
1.3.2
Ship loading problem
A small cargo company uses a container ship with a capacity to carry at most r freight containers. The ship
sails on a route with several stops at ports in between. At these ports, containers may be unloaded and new
containers may be loaded. At each port, there is a maximum amount bij of containers which is waiting to be
shipped from port i to port j > i (ports are numbered and visited in order from 1 to n). Let fij denote the
income per container shipped from port i to port j. The goal of the cargo company is to plan how much to
load at each port in order to maximize the income while never exceeding the total capacity of the ship.
Figure 4 shows the formulation of the problem as a minimum cost flow problem. The network has two
types of nodes:
P
• port nodes vi for 1 6 i 6 n with balance b(vi ) = − j<i bji
• shipment nodes vij for 1 6 i < j 6 n with balance b(vij ) = bij ; intuitively, vij represents the store of
containers waiting to be shipped from i to j: there will either be shipped by the cargo company or by
other means
The networks also has three types of arcs:
• shipped by company arcs e = (vij , vi ) with capacity u(e) = ∞ and cost c(e) = −fij ; intuitively, the flow
on e represents the amount of containers shipped by the company from i to j: the income per container
shipped can be seen as negative cost
• shipped by other means arcs e = (vij , vj ) with capacity u(e) = ∞ and cost c(e) = 0; intuitively, the
flow on e represents the amount of containers which are not shipped by the company from i to j, the
income per container (and thus the cost) is 0
• route arcs e = (vi, vi+1 ) with capacity u(e) = r and cost c(e) = 0; intuitively, the flow on e represents
the total amount of containers shipped from ports vi to vj by the company: it must not exceed the
capacity of the ship
Then the minimum cost flow exactly describes a loading of the ship which maximizes the income of the
company.
2
Mathematical properties
We study mathematical properties of network flows, in order to derive some theoretical as well as practical
results. We are interested in two kinds of results: simplifications of the problem (zero lower bounds, positive
costs), and necessary and sufficient conditions (feasible flow, minimum cost flow).
5
We will always assume from now on that the network contains no directed cycle of negative cost and
infinite capacity; indeed if there is such a cycle, then the minimum cost flow is undefined since we can build
flow of arbitrary small value.
We also define the following notations. If P is directed path or cycle, δe (P ) = 1 if e ∈ P and δe (P ) = 0
otherwise. Similarly δv (P ) = 1 if v is on P . If P is a directed path, α(P ) will refer to the starting node of P
and β(P ) to the ending node. And if G is graph, W is the set of all directed cycles of G and P is the set of all
directed paths of G. Furthermore, n will refer to the number of nodes and m to the number of arcs. We will
denote by U the maximum finite capacity and by B the maximum demand/supply: U = maxe {u(e) | u(e) <
∞} and B = maxv |b(v)|. We will also refer to C as the maximum cost: C = maxe |c(e)|. Finally, if e = (u, v)
is an edge, we denote by ee the reversed edge: ee = (v, u).
2.1
Flow decomposition
When formulating network flow problems, we can adopt either one of two equivalent modelling approaches.
We can define flows on arcs as usual or define flows on paths and cycles. Figure 5 shows those two models
on a example.
Definition 2.1.1 (Cycle and path flow). A cycle and path flow f is a flow fb defined by:
X
X
fb(e) =
δe (P )f (P ) +
δe (W )f (W )
P ∈P
W ∈W
P
We define the cost of a path or a cycle as c(P ) = e∈P c(e) and c(W ) =
a cycle and path flow as
X
X
f (P )c(P ) +
f (W )c(W )
P ∈P
P
e∈W
c(e). We define the cost of
W ∈W
This definition shows that each cycle and path flow determines arcs flows uniquely. The following theorem
shows that conversely, we can decompose any flow into a cycle and path flow.
x
f (e)
2
6
2
4 units
y
4
4
4
2 units
1
1
0
4
6
2
2
3
6
3
3
3
3
5
5
3 units
Figure 5: Two ways of representing flows in a network: using arc flows; or using path and cycle flows
Lemma 2.1.2. The cost of a cycle and path flow f equals the cost of the flow fb.
Proof.
!
X
fb(e)c(e) =
e∈A
=
X
X
e∈A
P ∈P
X
P ∈P
=
X
P ∈P
f (P )
δe (P )f (P ) +
X
δe (W )f (W ) c(e)
W ∈W
!
X
δe (P )c(e)
W ∈W
e∈A
f (P )c(P ) +
+
!
X
X
f (W )c(W )
W ∈W
6
f (W )
X
e∈A
δe (W )c(e)
Theorem 2.1.3 (Flow decomposition). Every flow f ◦ can be represented as a cycle and path f such that
fb = f ◦ with the following two properties:
• Every directed path with nonzero flow connects a supply node to a demand node
• At most n + m paths and cycles have nonzero flow; out of these, at most m cycles have nonzero flow.
Proof. We will give an algorithmic proof of the result. Suppose that v0 is a supply node. Then there is some
arc e = (v0 , v1 ) with nonzero flow. If v is a demand node, we stop with path P = (v0 , v1 ); otherwise the mass
balance constraint at v1 imposes that another arc (v1 , v2 ) has nonzero flow. We repeat this argument until
we either reach a demand node or encounter a previously visited node. One of these two cases will occur
within n steps. In the former case we obtain a directed path P from a supply node to a demand node and
in the latter case we obtain a directed cycle W . In either case, the path or the cycle only consists of nonzero
flow arcs.
If we obtain a directed path P from v0 to vk , we set f (P ) = min{b(v0 ), −b(vk ), mine∈P f (e)} and redefine
b(v0 ) = b(v0 ) − f (P ), b(vk ) = b(vk ) + f (P ) and f (e) = f (e) − f (P ) for every e ∈ P . If we obtain a directed
cycle W , we set f (W ) = mine∈W f (e) and redefine f (e) = f (e) − f (P ) for every e ∈ W .
We repeat this process with the redefined problem until there are no more supply nodes, that is b(v) = 0
for all v ∈ V . Then we select any node with one outgoing edge with nonzero flow as a starting point and
repeat the procedure. Since b(v) = 0 for all v ∈ V , we will go back to the starting point and find a directed
cycle as before. We finish when f = 0 for the redefined problem.
Clearly, the original flow is the sum of flows on the paths and cycles we found. Now notice that each time
we find a directed path, we redefine the demand or the supply to 0 or the flow on some arc to 0; similarly,
each time we find a directed cycle, we redefine the flow on some arc to 0. As a consequence, we will terminate
after identifying at most n + m cycles and paths and identify at most m cycles.
2.2
Removing nonzero lower bounds
In the most general statement of the problem, we allow arcs to have a nonzero lower bound l(e) the arc flow
f (e). Although this is useful in some problems, it is usually simpler to assume that l(e) = 0 in the proofs or
the algorithms.
To achieve this transformation, we consider another minimum cost flow problem with the same nodes
and arcs. We replace the flow f (e) in the original problem by f 0 (e) + l(e) where f 0 (e) is the flow in the
new problem. The flow bound constraint then becomes 0 6 f 0 (e) 6 u(e) − l(e). Making this substitution
in the balance constraints for e = (u, v) decreases b(u) by l(u)Pand increases b(v) by l(v). Finally, this
transformation decreases the objective function by the constant e∈A l(e)c(e).
Figure 6 illustrates this transformation on an arc. Then we can solve the original problem by solving the
new problem and transforming the solution back to this original setting.
b(u)
u
2.3
f (e)
(c(e), u(e))
b(v)
v
b(u) − l(e)
u
f ′ (e)
b(v) + l(e)
(c(e), u(e) − l(e))
v
Figure
6:
Removing
nonzero
lower
bounds
(f (e) = f 0 (e) + l(e))
Removing infinite capacities
In many, if not most, minimum cost flow problem, there are some uncapacitated arcs, i.e. arcs with u(e) = ∞;
such arcs are especially useful to simplify the problem formulation. However, in some transformations and in
many algorithms, it is not desirable to have uncapacitated arcs since we do some arithmetic with capacities
and flows. If an arc e has an infinite capacity, we wish to find a finite bound on the arc’s flow. Hopeful, we
usually know in advance an upper bound on the arc’s flow; for example in the ship loading problem, the flow
can’t be greater than the capacity of the ship and the shipments waiting in some ports. In the general case,
the following result provides an upper bound for such arcs.
7
Lemma 2.3.1. There exists f ◦ , a minimum cost flow such that f ◦ 6
X
v∈V
|b(v)| +
X
u(e)
e∈A
u(e)<∞
Proof. We already explained how we can transform the network in order to have zero lower bounds. We can
thus assume that l(e) = 0, for all e ∈ A. Now apply theorem 2.1.3 to f ◦ : we get a cycle and path flow f such
that fb = f ◦ . Thus if e0 ∈ A,
X
X
f ◦ (e0 ) =
δe0 (P )f (P ) +
δe0 (W )f (W )
P ∈P
W ∈W
Now pick a cycle W such that f (W ) > 0. Then we must have c(W ) 6 0 since f ◦ is a minimum cost flow;
otherwise we can set f (W ) = 0 and because we have zero lower bounds, we obtain a feasible flow of strictly
lesser cost than f ◦ . Furthermore, if c(W ) = 0 then we can set f (W ) = 0 and obtain a feasible flow of the
same cost. So we can assume that if f (W ) > 0 then c(W ) < 0.
But remember that f ◦ is a minimum cost flow so if there is a cycle W such that c(W ) < 0 then at least
one edge eW of W must be saturated, that is f ◦ (eW ) = u(eW ) < ∞ otherwise we could push some more flow
on W and get a better flow. Let W + = {W | f (W ) > 0} and AW = {eW , W ∈ W + }. Then we have:
X
X
X
X
X
δe0 (W )f (W ) 6
f (W ) 6
f ◦ (e) 6
u(e) 6
u(e)
W ∈W
W ∈W +
e∈AW
e∈AW
e∈A
u(e)<∞
Now rewrite the mass balance constraint with cycle and path costs for a node v:
X
X
X
X
b(v) =
f (e) −
f (e) =
f (P ) −
f (P )
e∈N + (v)
P ∈P
α(P )=v
e∈N − (v)
P ∈P
β(P )=v
But remember that from theorem 2.1.3, all directed path go from supply nodes to demand nodes, so for each
v, either {P | α(P ) = v} = ∅ or {P | β(P ) = v} = ∅. Thus we have:
X
X
|b(v)| =
f (P ) +
f (P )
P ∈P
α(P )=v
P ∈P
β(P )=v
So finally:
X
δe0 (P )f (P ) 6
P ∈P
X
f (P ) 6
P ∈P
X X
v∈V
P ∈P
α(P )=v
f (P ) 6
X
v∈V
|b(v)|
Putting things together proves the result.
2.4
Removing negative costs
In many problem formulations it is convenient to work with negative costs; indeed minimizing a negative cost
quantity is equivalent to maximizing a positive profit. However, it is much more convenient to only allow
nonnegative costs in the proofs and algorithms. In this transformation, we need to handle uncapacitated arcs
carefully. Thus, we rely on lemma 2.3.1 to ensure that all capacities are finite.
In this transformation we consider another minimum cost flow problem with the same nodes. All the arcs
with nonnegative cost are kept in the new problem and for these f (e) = f 0 (e) where f 0 is the flow in the
new network. On the contrary, for each arc e = (u, v) with negative cost, we add an edge e0 = ee in the new
problem, with cost c0 (e0 ) = −c(e) and we replace the flow f (e) in the original network by u(e) − f 0 (e). To
account for the change in balance, we decrease b0 (u) by u(e) and increase b0 (v) by u(e).
Figure 7 illustrates this transformation on an arc. Then we can solve the original problem by solving the
new problem and transforming the solution back to this original setting. The intuition of this transformation
is that instead of sending f (e) units of flows and paying c(e) each, we beginning by sending u(e) and then
f 0 (e) account for the diminution of the flow. Then decreasing the flow costs −c(e). And we change the
demand/supply of the nodes to make everything consistent.
8
b(u)
u
2.5
b(v)
f (e)
(c(e), u(e))
v
b(u) − u(e)
u
f ′ (e)
b(v) + u(e)
Figure 7: Working with positive costs (f (e) = u(e) −
f 0 (e))
v
(−c(e), u(e))
Making the network strongly connected
When working with reduced costs, it will be easier to assume that there is path between each pair of nodes;
that is, that the network is strongly connected. And even stronger, we will want that there is an uncapacitated
path between each pair of nodes. Since this hypothesis is hardly verified in general, we need to explain how
to transform the network to make it strongly connected. To do so, we will assume that the network has finite
capacities, which is always possible thanks to transformation 2.3. Notice that we will add arcs with infinite
capacities in this transformation, so one will eventually have to make these finite but these transformations
will not “loop”.
s
v
s
MC
∞
∞
MC
v
Figure 8: Making the network strongly connected (c(e) = M C, u(e) = ∞)
P
P
Now define M = {|b(v)|, v ∈ V } + {u(e), e ∈ A | u(e) < ∞}. From 2.3.1, we know that there is a
minimum cost flow f ∗ which satisfies ∀e ∈ A, f ∗ (e) 6 M . Now, fix a node s ∈ V and for each v ∈ V , add
an arc e = (s, v) and an arc e0 = (v, s) with cost c(e) = c(e0 ) = 1 + M C and capacity u(e) = u(e0 ) = ∞.
After this transformation, the network clearly satisfies the wanted property: there is an uncapacitated path
between each pair of nodes. Figure 8 illustrates the transformation on an example. But what about the
minimum cost flows ? The following lemma shows that they remain the same.
Lemma 2.5.1. The minimum cost flows remain the same after the transformation.
Proof. Let A0 be the set of arcs after the transformation. Let f ◦ be a minimum cost flow. Then by definition
of f ◦ and f ∗ we have:
X
X
f ◦ (e)c(e) =
f ∗ (e)c(e)
e∈A0
e∈A
0
Now suppose that there is an arc e0 ∈ A \ A such that f ◦ (e0 ) > 0, i.e. there is a minimum cost flow which
use one of the new arcs. Then we must have (because all the data are integral):
X
X
f ◦ (e)c(e) > f ◦ (e0 )c(e0 ) > c(e0 ) = 1 + M C > M C >
f (e)c(e)
e∈A0
e∈A
◦
◦
Which contradicts the optimality of f . Thus f does not use any new arc.
2.6
Residual network
When implementing network flow algorithms, we are not thinking in terms of absolute flow but rather in
terms of incremental flow. To do so, we define the residual network, that gives information about how we
can push or extract some flow into or from the network.
The idea behind the residual network is the following. Suppose that an edge e carries f ◦ (e) units of flow,
then we can still push up to u(e) − f ◦ (e) units of flow on this edge. But we can also extract up to f ◦ (e) units
of flow from it; to do so we can push up to f ◦ (e) units of flow on ee, which amounts to cancelling the existing
flow on e. Finally notice that pushing a unit of flow on e costs c(e) and that pushing one unit of flow on ee
costs −c(e); since we are removing some flow.
9
Formally, the residual network with respect to a given flow f ◦ is a network with the same nodes defined
as follows. For each arc e in the original network, we add an arc e with cost c(e) and residual capacity
r(e) = u(e) − f ◦ (e); and the arc ee with cost −c(e) and residual capacity f ◦ (e). The residual network consists
only of arcs with a positive residual capacity. Thus we delete arcs with a zero capacity. We denote by
G(f ◦ ) this network.
Figure 9 illustrates the construction for an arc. Notice that the residual graph may have parallel arcs
even though the original network doesn’t have parallel ones.
b(u)
u
b(v)
f ◦ (e)
(c(e), u(e))
0
u
v
(c(e), u(e) − f ◦ (e))
(−c(e), f ◦ (e))
0
v
Figure 9: Construction of
the residual network
The main interest of the residual network is that it makes it easier to state properties while preserving a
one-to-one correspondence between feasible flows in the two networks that preserves the cost of the solutions.
We state a more precise version of this result in the following theorem.
Theorem 2.6.1. Let f ◦ be a feasible flow of a network N and G = G(f ◦ ). Then a flow f is feasible in N if
and only if the corresponding flow f 0 defined by f 0 (e) − f 0 (e
e) = f (e) − f ◦ (e) and f 0 (e)f 0 (e
e) = 0 is a feasible
flow in G. Furthermore,
X
X
X
c(e)f (e) =
c0 (e)f 0 (e) +
c(e)f ◦ (e)
e∈A
e∈AG
e∈A
where c is the cost in N and c0 the cost in G.
Proof. First notice that the flow f 0 is well-defined. Indeed, the condition f 0 (e)f 0 (e
e) = 0 implies that either
f 0 (e) = 0 or f 0 (e
e) = 0. But if f 0 (e) = 0 then f 0 (e
e) = f ◦ (e) − f (e); and if f 0 (e
e) = 0 then f 0 (e) = f (e) − f ◦ (e).
Thus, it only depends on the sign of f (e) − f ◦ (e).
⇒ First assume that f is a feasible flow and define f 0 as above. Then f 0 must satisfy the bound constraints,
there are two cases to check:
• if f (e) > f ◦ (e) then 0 6 f 0 (e) = f (e) − f ◦ (e) 6 u(e) − f ◦ (e) = r(e) because f (e) 6 u(e). Furthermore
f 0 (e
e) = 0 so it satisfies the bound constraint
• if f (e) 6 f ◦ (e) then 0 6 f 0 (e
e) = f ◦ (e) − f (e) 6 f ◦ (e) = r(e
e). Furthermore, f 0 (e) = 0 so it satisfies the
bound constraint
Then, we can check that the mass balance constraint is satisfied with a simple calculus:
X
X
X
X
f 0 (e) −
f 0 (e) =
(f 0 (e) − f 0 (e
e)) −
(f 0 (e) − f 0 (e
e))
+
e∈NG
(v)
−
e∈NG
(v)
=
e∈N + (v)
e∈N − (v)
X
X
(f (e) − f ◦ (e)) −
(f (e) − f ◦ (e)) = b(v) − b(v) = 0
e∈N − (v)
e∈N + (v)
Finally, we can link the relationship on the cost as follows (and then summing over the arcs).
c0 (e)f 0 (e) + c0 (e
e)f 0 (e
e) = c(e)(f 0 (e) − f 0 (e
e)) = c(e)(f (e) − f ◦ (e)) = c(e)f (e) − c(e)f ◦ (e)
⇐ Now assume that f 0 is a feasible flow. Then f (e) = (f 0 (e) − f 0 (e
e)) + f ◦ (e) and f must satisfy the bound
constraints, there are two cases to check:
• if f 0 (e) = 0 since 0 6 f 0 (e
e) 6 r(e
e) = f ◦ (e), we have 0 6 f (e) 6 f ◦ (e) 6 u(e)
10
• if f 0 (e
e) = 0 since 0 6 f 0 (e) 6 r(e) = u(e) − f ◦ (e), we have 0 6 f ◦ (e) 6 f (e) 6 u(e)
Then we can easily check the mass balance constraint is satisfied:
X
X
X
X
f (e) −
f (e) =
(f 0 (e) − f 0 (e
e) + f ◦ (e)) −
(f 0 (e) − f 0 (e
e) + f ◦ (e))
e∈N + (v)
e∈N − (v)
e∈N − (v)
e∈N + (v)
=
X
f ◦ (e) −
e∈N + (v)
X
f ◦ (e) +
e∈N − (v)
X
f 0 (e) −
+
e∈NG
(v)
X
f 0 (e)
−
e∈NG
(v)
= b(v) + 0 = b(v)
Finally, the calculus about cost holds in both cases, so there is no need to redo it.
2.7
Negative cycle optimality condition
Thanks to the properties of the residual network, we are now able to state a first optimality condition for the
minimum flow problem. As we will see, this condition give rise to very simple (but inefficient algorithm) for
solving the problem.
Theorem 2.7.1. A feasible flow f ∗ is an optimal solution of the minimum cost flow problem if and only if
G(f ∗ ) contains no negative cost directed cycle.
Proof. ¬ ⇐ ¬ Assume that f ∗ is an feasible flow and that G(f ∗ ) contains a negative cost directed cycle W
of cost c0 (W ) < 0 and maximum flow f 0 (W ) > 0 on it. Then f ∗ cannot be optimal because we can augment
it along W and decrease its value by −c0 (W )f 0 (W ). Formally, consider the residual flow f 0 (e) = δe (W )f (W ).
Then f 0 is a feasible flow in G(f ∗ ) by hypothesis and by applying theorem 2.6.1 we get a flow feasible f
which is strictly better than f ∗ and thus yields a contradiction:
X
X
X
c(e)f (e) −
c(e)f ∗ (e) =
c0 (e)f 0 (e) = c0 (W )f 0 (W ) < 0
e∈A
e∈A
e∈AG
¬ ⇒ ¬ Assume that f ∗ is a feasible flow and that G(f ∗ ) contains no negative cost directed cycle. Let f ◦
be any feasible flow. We will show that f ∗ has a lesser cost than f ◦ . Apply theorem 2.6.1 to f ◦ in G(f ∗ ).
Then we get a residual flow f 0 . Now apply the flow decomposition theorem (2.1.3) to f 0 in G(f ∗ ). We get a
cycle and path flow x such that x
b = f 0 . But recall that the mass balance constraint in G(f ∗ ) implies that
all nodes are transshipment nodes. Thus x is only a cycle flow. But since G(f ∗ ) contains no negative cost
directed cycle, the cost of any cycle must be nonnegative and thus x has a nonnegative cost:
X
c0 (W )x(W ) > 0
W ∈W(G(f ∗ ))
Now, using the cost relationship of a cycle and path flow (2.1.2) we get that:
X
X
X
c0 (e)f 0 (e) =
c0 (e)b
x(e) =
c0 (W )x(W ) > 0
e∈AG
W ∈W(G(f ∗ ))
e∈AG
And finally, using the cost relationship of the residual network network, we get that:
X
X
X
c(e)f ◦ (e) −
c(e)f ∗ (e) =
c0 (e)f 0 (e) > 0
e∈A
e∈A
e∈AG
Thus, f ∗ is an optimal flow.
11
2.8
Reduced cost
In many minimum cost flow algorithms, we measure the cost of an arc related to some cost associated with
the nodes. These costs are typical intermediate data of the algorithm. Assume that we associate to each
node v ∈ V a potential π(v). We then define the reduced cost with respect to π as cπ defined by:
cπ (e) = c(e) + π(v) − π(u)
if e = (u, v)
In those algorithms, the reduced cost replaces the cost, especially in the residual network. Thus, it is crucial
to understand the relationship between c and cπ , with respect to the minimum flow problem. We first relate
the objective function for c with the objective function for cπ :
Lemma 2.8.1. A minimum cost flow problem has the same optimal solutions if we replace the cost c by cπ .
Furthermore, for any feasible flow f we have:
X
X
X
f (e)c(e) −
f (e)cπ (e) =
π(v)b(v)
e∈A
e∈A
v∈V
Proof. First, we establish the relationship on the cost:
X
X
X
f (e)cπ (e) =
f (e)c(e) +
e∈A
e∈A
=
X
e=(u,v)∈A
f (e)c(e) +
X
e∈A
v∈V
X
X
π(v)
f (e)π(v) −
X
e∈N − (v)
X
f (e)π(u)
e=(u,v)∈A
f (e) −
X
u∈V
π(u)
X

=
=
f (e)c(e) +
e∈A
v∈V
X
X
e∈A
f (e)c(e) −
f (e)
e∈N + (v)

X
π(v) 
e∈N − (v)
f (e) −
X
f (e)
e∈N + (v)
π(v)b(v)
v∈V
Thus the total decrease in the objective function does not depend on the flow. Therefore if f minimizes the
objective function for c, it minimizes the objective function for cπ .
We can also study the effect of the reduced cost on the cycle and paths flows that we introduced for the
flow decomposition:
Lemma 2.8.2. For any network flow with cost function c, for any potential function π we have:
P
P
1. For any directed cycle W , e∈W c(e) = e∈W cπ (e)
P
P
2. For any directed path from u to v, e∈P cπ (e) = e∈P c(e) + π(v) − π(u)
Proof. trivial
The introduction of reduced costs is related to another network flow problem, namely the shortest path
problem. In this context, the potential can seen as the potential shortest path. We now present a result to
link the shortest path problem to the use of potentials and give an intuition about minimum cost flow:
Lemma 2.8.3. Assume there is a source node s ∈ V such that every node v ∈ V is accessible from s. Let d
be a potential function and π = −d. Then if there is no negative cost cycle in the network, the following two
conditions are equivalent:
(i) d(v) is the shortest path distance from s to v with respect to c
(ii) d(s) = 0 and cπ (e) > 0
∀e ∈ A
12
Proof. We will implicitly use the fact that d is well-defined, that is, there is no negative cost cycle.
(i) ⇒ (ii) Let e = (x, v) ∈ A. Now write the fact that d(v) is the shortest path distance from s to v with
respect to c:
⇒
d(v) 6 d(x) + c(e)
c(e) + d(x) − d(v) > 0
⇒
cπ (e) = c−d (e) > 0
¬(i) ⇒ ¬(ii) There are two cases: either d(s) 6= 0 and in this case d trivially cannot be the shortest path
distance (since the shortest path distance from s to s must be 0). Either ∃e0 , cπ (e0 ) < 0. By writing
e0 = (x0 , v0 ) we get that:
cπ (e0 ) < 0
⇒
c(e0 ) + d(x0 ) − d(v0 ) < 0
⇒
d(x0 ) + c(e0 ) < d(v0 )
Which contradicts the fact that d is the shortest path distance function.
2.9
Reduced cost optimality condition
We will now try to merge the result about shortest path reduced cost (2.8.3) and negative cycle optimality
condition (2.7.1). To do so, we will have to assume that the network is strongly connected and there is an
“uncapacitated” path between each pair of vertices, a property which does not incur any loss of generality,
thanks to section 2.5. By “uncapacitated” we mean high enough, because we actually want all capacities to
be finite but want to be able to push any reasonable amount of flow.
The intuition of the reduced cost optimality condition is the following. We know that a flow is optimal if
there is no negative cost cycle in the residual network. We will try to reformulate the last part. If there is
no negative cost cycle in the residual network, it means that the shortest path distance is well defined with
respect to the cost function. On the contrary, if there is a negative cost cycle, the shortest path distance is
not well-defined. We will express the “well-definedness” of the shortest path in terms of reduced costs, thanks
to the property 2.8.3. This give rise to the following theorem.
Theorem 2.9.1 (Reduced cost optimality condition). A feasible flow f ∗ is an optimal solution of the minimum cost flow problem if and only if ∃π a potential function satisfying the reduced cost optimality condition
in the residual network G(f ∗ ):
cπ (e) > 0 ∀e ∈ A(G(f ∗ ))
Proof. ⇒ Assume that f ∗ is an optimal solution. By the negative cycle optimality condition theorem (2.7.1),
there is no negative cost cycle in G(f ∗ ). Thus the shortest path distance d(·) in G(f ∗ ) is well-defined. Now
apply lemma 2.8.3 to G(f ∗ ) and we get that the potential π = −d satisfy the reduced cost optimality
condition.
⇐ Assume that there exists a potential function π such that the reduced cost optimality condition is
satisfied. We will show that any cycle in G(f ∗ ) has nonnegative cost. Then by the negative cycle optimality
condition theorem (2.7.1), we will conclude that f ∗ is optimal. Let W be a directed cycle in G(f ∗ ). By
lemma 2.8.2, we can link the cost of W in G(f ∗ ) to its reduced cost and because all the reduced costs are
nonnegative, the cycle must have a nonnegative cost:
X
X
c(e) =
cπ (e) > 0
because cπ (e) > 0, ∀e ∈ A(G(f ∗ ))
e∈W
3
e∈W
Algorithms
In this section, we describe a simple algorithm to solve the minimum cost flow problem in a very inefficient
way. There are many more algorithms to solve this problem, some weakly polynomial time ones and even
some strongly polynomial time ones but we refer to [1] for more details.
In this section, we assume that all networks satisfy the following properties:
13
• The network has no directed cycle of negative cost and infinite capacity, as before
• All capacities, costs and demands/supplies are integral, as before
• All lower bounds are zero; we refer to 2.2 for more details
• All capacities are finite; we refer to 2.3 for more details
• All arcs costs are nonnegative; we refer to 2.4 for more details
• There is an “uncapacitated” path between each pair of vertices; we refer to 2.5 for more details
One should notice that these assumptions imply no loss of generality and do not change the complexity of
the algorithm.
3.1
Finding a feasible flow
A large class of minimum flow algorithms start by computing a feasible flow and then improve it until it
reaches optimality. Finding a feasible flow is not a trivial problem because of the mass balance constraint.
To determine a feasible flow, we will reduce the problem to a maximum flow problem.
Consider the network G with the same nodes and arcs as the original network, but without any costs.
Now add two special nodes s∗ and t∗ and add arcs as follows. For each v ∈ V , if b(v) > 0 then add an arc
(s∗ , v) with capacity b(v), if b(v) < 0 then add an arc (v, t∗ ) with capacity −b(v). Now solve a maximum
flow problem from s∗ to t∗ in the new network. If the maximum flow saturates all the source arcs, then the
original network has a feasible flow; otherwise it is infeasible. Furthermore, to obtain a feasible flow from the
maximum flow, we just need to forget about the arcs we added.
b(u) > 0
b(u) < 0
u
u
b(u
)
(u
s∗
u
−b
)
u
t∗
Figure 10: Transformation from the feasible flow to the maximum flow problem
Figure 10 illustrates the transformation on a node. The following lemma gives a mathematical basis to
this transformation. Algorithm 1 describes the pseudo-code of such an algorithm.
Lemma 3.1.1. The original network has a feasible flow if and only if every maximum flow of G saturates
all the source arcs. Furthermore, if f ∗ is a maximum flow of G. Then f defined by f = f|V is a feasible flow
of the original network.
Proof. trivial.
3.2
Cycle-cancelling algorithm
The negative cycle optimality condition stated in theorem 2.7.1 suggests a natural algorithm to find minimum
cost flow, which is commonly called the cycle-cancelling algorithm. This algorithm starts with a feasible flow
and tries to find a negative cost cycle in the residual graph. When no such cycle exists, we know we have an
optimal flow; otherwise, we augment the flow along the cycle, decrease the cost of the flow and repeat the
procedure. Algorithm 2 describes the pseudo-code of this an algorithm. Theorem 3.2.1 gives an upper bound
on complexity of this algorithm.
Theorem 3.2.1. Algorithm 2 has running time O(nm2 CU ).
Proof. Consider the following points:
14
Function 1: FeasibleFlow(N, b, u)
Data: A network N = (V, A); a demand/supply function b; a capacity function u
Result: A feasible flow or ∅
begin
V ← V ∪ {s∗ , t∗ }
foreach v ∈ V do
if b(v) > 0 then
A ← A ∪ {(s∗ , v)}
u(s∗ , v) ← b(v)
else if b(v) < 0 then
A ← A ∪ {(v, t∗ )}
u(v, t∗ ) ← −b(v)
f ←MaximumFlow(N, u)
if f saturates all source arcs then return f
else return ∅
Function 2: CycleCanceling(N, b, u)
Data: A network N = (V, A); a demand/supply function b; a capacity function u; a cost function c
Result: A minimum cost flow or ∅
begin
f ←FeasibleFlow(N, b, u)
if f = ∅ then return ∅
(G, c0 , r) ←ResidualGraph(N, c, u)
W ←NegativeCostCycle(G, c0 )
while W 6= ∅ do
δ ← min{r(e), e ∈ W }
f ←AugmentFlow(f, δ, W, c0 )
(G, c0 , r) ←ResidualGraph(N, c, u)
W ←NegativeCostCycle(G, c0 )
return f
• Finding a feasible flow reduces to finding a maximum flow which can be done in O(n2 m) time (using a
variant of Ford and Fulkerson algorithm which augments along shortest paths, see [2] or [1])
• The
P initial flow has cost at most mCU : since f (e) 6 U and c(e) 6 C, c(e)f (e) 6 CU and thus
e∈A c(e)f (e) 6 mCU
• The minimum cost flow has cost at least 0 since all costs are nonnegative
• Each iteration decreases the flow cost by at least 1 since all the data are integral
• Finding a negative cost directed cycle in a graph can be done in O(nm) time (using the Bellman and
Ford algorithm, see [2] or [1])
Thus, the algorithm executes at most O(mCU ) times a O(nm) algorithm, plus a O(n2 m) algorithm. Which
gives a total O(nm2 CU ) running time.
Notice that this algorithm only has a pseudo-polynomial running time since it depends on C and U instead
of log C and log U .
3.3
Successive shortest path algorithm
The cycle-cancelling algorithm works by finding a feasible flow and decreasing the total cost until a certain
condition is satisfied and implies optimality (namely the negative cycle condition). On the contrary, we will
15
now present an algorithm based on the reduced cost optimality condition which works the other way around.
That is, it will maintain a pseudoflow which always satisfy the optimality condition and will push flow on
it until it becomes a real flow. This pseudoflow will satisfy the capacity constraints but will violate the
mass-balance constraint.
Definition 3.3.1 (Pseudoflow). A pseudoflow is a function f : A → R+ satisfying only the capacity constraints. If f is a pseudoflow, we define the imbalance at a node v as:
X
X
e(v) = b(v) +
f (e) −
f (e)
∀v ∈ V
e∈N − (v)
e∈N + (v)
If e(v) > 0, we say that e(v) is the excess of v; and if e(v) < 0, we say that −e(v) is the deficit of v; otherwise
v is said to be balanced.
Notice that since the total balance of the network is 0, if the network has an excess node, it must have
a deficit node. We will now apply the reduced cost optimality condition to the concept of pseudoflow to
see how we can derive an algorithm. The idea is that given a pseudoflow and potential function satisfying
the reduced cost optimality condition, we will push some flow into the network and build a new potential
function satisfying the reduced cost optimality condition.
Lemma 3.3.2. Assume that a pseudoflow f satisfies the reduced cost optimality condition for some potential
π. Let d(·) be the shortest path distance function from a node s ∈ V to all other nodes in the residual graph
G(f ) with respect to the reduced cost cπ . Then, the following two properties hold:
(i) f satisfies the reduced cost optimality condition with respect to the potential π 0 = π − d
0
(ii) for each arc e in a shortest path from s to any other node, cπ (e) = 0
Proof. Since d(·) is the shortest path distance function in G(f ) with respect to cost cπ , we can apply lemma
2.8.3 which yields that for each arc e = (u, v) ∈ A(G(f )):
−d
(cπ )
(e) > 0
⇒
cπ (e) + (−d)(v) − (−d)(u) > 0
⇒
cπ (e) > 0
⇒
c(e) + (π(v) − d(v)) − (π(u) − d(u)) > 0
0
Now if e = (u, v) is in a shortest path from s to any node with respect to cost cπ , we have by definition:
d(v) = d(u) + cπ (e)
⇒
0
cπ (e) = 0
As explained before, this lemma implies that we can augment the pseudoflow along a shortest path and still
get a pseudoflow satisfying the reduced cost optimality condition. The next lemma formalizes this intuition.
Recall that r(e) is the residual capacity of e in the residual graph G(f ).
Lemma 3.3.3. Assume that a pseudoflow f satisfies the reduced cost optimality condition for some potential
π. Let d(·) be the shortest path distance function from a node s ∈ V to all other nodes in the residual graph
G(f ) with respect to the reduced cost cπ . Let P be a shortest path from s to any node vP .
Define µ = min(e(s), −e(vP ), mine∈P r(e)). Define f 0 by f 0 (e) = f (e) + δe (P )µ. Then f 0 is a pseudoflow that
satisfies the reduced cost optimality condition for π 0 = π − d.
Proof. The fact that f 0 is a pseudoflow is a direct consequence of the choice of µ and the fact that P is a
0
path in G(f ). Now, from lemma 3.3.2 we know that for each arc e ∈ P , since P is a shortest path, cπ (e) = 0.
Thus e satisfies the reduced cost optimality condition with respect to π 0 . But recall that augmenting the flow
0
0
along e might add the arc ee to the residual graph. Hopefully, cπ (e
e) = −cπ (e) = 0. Thus, if it exists, ee also
satisfies the reduced cost optimality condition with respect to π 0 .
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We now have all the ingredients to describe the successive shortest path algorithm. Lemma 3.3.3 gives the
key idea of the algorithm. The potential will be at the centre of the algorithm; we will maintain at the same
time a pseudoflow f and a potential π. At each step of the algorithm, we will augment f along a shortest
path with respect to the reduced cost cπ . By doing so, we will reduce the excesses and deficits in the network
until all nodes are balanced. At this point, the pseudoflow will become a flow which satisfies the reduced cost
optimality condition, and from theorem 2.9.1 we will know that this flow is optimal. Algorithm 3 describes
the pseudo-code of this algorithm. Theorem 3.3.4 gives an upper bound on complexity of this algorithm.
Function 3: SuccessiveShortestPath(N, b, u)
Data: A network N = (V, A); a demand/supply function b; a capacity function u; a cost function c
Result: A minimum cost flow or ∅
begin
f (e) ← 0 for all e ∈ A
π(e) ← 0 for all e ∈ A
E ← {v ∈ V | e(v) > 0}
D ← {v ∈ V | e(v) < 0}
while E 6= ∅ do
pick s ∈ E and v ∈ D
(G, c0 , r) ←ResidualGraph(N, cπ , u)
d ←ShortestPathDistance(G, cπ , s)
P ←ShortestPath(G, cπ , d, s, v)
µ ← min(e(s), −e(v), mine∈P r(e))
f ←AugmentFlow(f, µ, P, c0 )
π ←π−d
E ← {v ∈ V | e(v) > 0}
D ← {v ∈ V | e(v) < 0}
return f
Theorem 3.3.4. Algorithm 3 has running time O(nm log(n)B).
Proof. Consider the following points:
• Initialization steps are negligible compared to the cost of the loop body
P
• The initial pseudoflow has a total excess (i.e v∈E e(v)) of at most nB.
• The minimum cost flow has a total excess of 0
• Each iteration decreases the excess of some node (and thus the total excess) by at least 1 since all the
data are integral
• Finding the shortest path distance in a directed graph with nonnegative costs (since cπ satisfies the
reduced cost optimality condition) can be done in O(m log(n)) time (using the Dijkstra’s algorithm
with a binary heap, see [2] or [1])
Thus, the algorithm executes at most O(nB) times a O(m log(n)) algorithm. Which gives a total O(nm log(n)B)
running time.
Notice that this algorithm only has a pseudo-polynomial running time since it depends on B instead of log B.
However, this is already much better that the cycle cancelling algorithm in general. Furthermore, using
some capacity scaling and doing some very small modifications to the algorithm, it is possible to obtain a
weakly-polynomial algorithm. The idea is that instead of augmenting along a shortest path between any pair
of vertices, we first augment the flow between vertices with high excess and deficit.
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4
Conclusion
We saw a mathematical formulation of the minimum cost flow problem and we saw that it is a natural
generalisation of the maximum flow problem and the shortest path problem. We also gave the intuition
that a number of real-life problems can easily be stated in terms of minimum cost flow. We then studied
some properties of the minimum cost flows in a network. In particular, we gave two different optimality
conditions which naturally require the introduction of the residual network and reduced costs. Finally, we
presented two pseudo-polynomial time algorithms based on the conditions developed previously. Although
some weakly-polynomial and strongly-polynomial algorithms exist to solve this problem, they require more
material. In particular, it should be noted that it is possible to obtain such algorithms with only relatively
simple modifications to the one we presented. Finally, as the minimum cost flow is a generalisation of the
maximum flow problem, there are further generalisation of the problem like convex flows, generalized flows
or multicommodity flows.
References
[1] Ravindra K. Ahuja, . Magnanti, Thomas L., and James B. Orlin. Network flows : theory, algorithms,
and applications / Ravindra K. Ahuja, Thomas L. Magnanti, James B. Orlin. Prentice Hall, Englewood
Cliffs, N.J. :, 1993.
[2] Thomas H. Cormen, Clifford Stein, Ronald L. Rivest, and Charles E. Leiserson. Introduction to Algorithms. McGraw-Hill Higher Education, 2nd edition, 2001.
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